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Integral question - $$\int\frac{\sqrt{\tan(x)}}{{\cos^2(x)}}dx$$
I see that $\frac{1}{\cos^2(x)}$ is the derivative of $\tan(x)$ so I set $t = \tan(x)$? or the whole square?
Thanks!

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1  
$\tan'x=\sec^2x$ –  1015 May 1 '13 at 20:51
    
Just $\tan(x)$... –  David Mitra May 1 '13 at 20:52

2 Answers 2

up vote 3 down vote accepted

It is easier to set just $\tan(x) = t$. If you set $\tan(x) = t$, we get that $\sec^2(x) dx = dt$, i.e., $\dfrac{dx}{\cos^2(x)} = dt$. Hence, we get $$\int \dfrac{\sqrt{\tan(x)}}{\cos^2(x)}dx = \int \sqrt{t} dt$$ I trust you can take it from here.

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Hint: for a derivable function $\,f(x)\,$ , we have that

$$\int\sqrt{f(x)}\;f'(x)\;dx=\frac23f(x)\sqrt{f(x)} + C$$

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